Boson-pair realization of SU(1,1) ($K_{+}=a^{†}{b}^{†},K_{-}=ab,K_3=(N_a+N_b+1)/2$)

Pair creation and annihilation in two independent oscillator modes obey $[K_{-},K_{+}]=2K_3$ and $[K_3,K_{\pm }]=\pm K_{\pm }$. The Hermitian components $K_1=(K_{+}+K_{-})/2$ and $K_2=(K_{+}-K_{-})/(2i)$ satisfy $[K_1,K_2]=-i{K}_3$. Multiplying all three Hermitian components by $i$ gives the real Lie bracket signs of SU(1,1). Both occupations change equally, so their difference is conserved. This realizes the algebra used in two-mode squeezed states.

The identities $K_{+}{K}_{-}=N_a{N}_b$ and $K_{-}{K}_{+}=(N_a+1)(N_b+1)$ give the displayed quadratic Casimir operator. It commutes with all generators because the number difference does. A fixed number-difference sector therefore has a fixed Casimir value even though pair creation changes its total occupation.